9 votes
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Multiplication by a Haar random unitary two times

There is an explicit formula for the integral with respect to the Haar measure of any polynomial in the entries of a unitary and its conjugate, due to Collins and Śniady: Benoît Collins and Piotr ...
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  • 4,568
8 votes
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Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

This answer won't actually give you a bound, but will provide some information that may help you in your search. You may be able to find an answer in the random matrix theory literature if you ...
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  • 4,568
8 votes

What is a Haar random quantum state?

Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$...
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6 votes

What is the expectation value ${\Bbb E}[\langle\psi,O\psi\rangle]$ over the Haar distribution?

Since a Haar-random $\lvert\psi\rangle=U\lvert0\rangle$ for a Haar-random $U$, your expectation value equals $$ \langle 0 \rvert \Big[\int \mathrm d U\, UOU^\dagger\Big]\lvert0\rangle\ . $$ The ...
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6 votes

How to show that the integral over all Haar states vanishes: $\int|\psi\rangle\,{\rm d}\psi = 0 $?

Yes, we can show this using the unitary invariance of the Haar measure on states. In more detail, we have $$ U \int |\psi\rangle\, \mathrm{d}\psi = \int U|\psi\rangle\, \mathrm{d}\psi = \int |\psi\...
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6 votes
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How close or far apart are the distributions generated by two Haar random states?

Since the Haar-measure is unitarily invariant, the $\mathbf{D}_\psi$ that we obtain will be independent of $\psi$. In fact, the $\mathbf{D}_\psi$ obtained from measuring $\psi$ with respect to any ...
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6 votes

Expected value of a Haar random quantum state multiplied by a unitary

I'm writing an alternate proof because it uses some interesting tools, computes the value of these expressions, and gives some insights into how we can interpret the quantities in consideration. The ...
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5 votes
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Confusion about the output distribution of Haar random quantum states

The two facts are connected in that they both arise as a result of rotational invariance of the Haar measure. We will derive them in the case of large $n$ since this is when the Porter-Thomas ...
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5 votes
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Question regarding integration of Haar random state

The issue that easily leads to confusion is the dual role played by output bitstring probability. It enters the computation of the average in two ways. On one hand, it determines how often one sees ...
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4 votes

Is there a lower bound on the average diamond norm of two uniformly random unitaries U1 and U1 of dimension D that are sampled from haar measure?

The result you're looking for is effectively Proposition 19 of the paper: Almost all quantum channels are equidistant; which I'm rewriting here for convenience: Let $U, V \in \mathcal{U}(d)$ be two ...
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4 votes
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Expected value of a Haar random quantum state multiplied by a unitary

With the chosen structure of $ U $, i think it's even possible to prove the stronger statement: $$ \langle z| \rho|z \rangle = \langle z| \sigma_\rho|z \rangle, \hspace{0.2em} \text{where} \hspace{0....
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  • 1,326
4 votes

Is the Haar measure invariant under conjugation?

I will answer this question in a more general context. You might know that Haar's theorem tells you that on any locally compact group $G$, there is a unique left-invariant (Borel) measure $\mu$, up to ...
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4 votes

Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Suppose that was not the case. Then taking the first column of a uniformly random unitary matrix gives you a nonuniformly random state. That means that there is some state, call it $|\psi\rangle$, ...
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4 votes

Is the column vector of a uniformly sampled random unitary matrix a uniformly sampled random state vector?

Yes. A uniformly (Haar random) sampled state vector $|\psi\rangle$ is characterized by the fact that the probability measure is invariant under any $U$, i.e., colloquially, $U|\psi\rangle$ is just as ...
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4 votes
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At what depth and for what architecture are random quantum circuits $1$-designs?

To study unitary $t$-designs, we define the moment operator with respect to a probability measure $\nu$ as $$ M_t(\nu) := \int_{U(d)} U^{\otimes t} (\cdot) (U^{\otimes t})^\dagger d\nu(U) \simeq \int_{...
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4 votes
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Anticoncentration for two independent random quantum circuits in parallel

I believe so (caveat: this is not something I've every thought about before). I'm going to rewrite the $p_x$ from your question as $p_{xy}$. So, we have $$ p_{xy}=|\langle x|U_1|0^n\rangle|^2\ |\...
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  • 47.5k
3 votes
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Random quantum states and Schur-Weyl duality

Note that the quoted relation $$ \bar M_i = \sum_\lambda a_\lambda P_\lambda, $$ only holds if the $M_i$ also commute with the representation of the symmetric group! Otherwise this can obviously not ...
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3 votes

Spoofing XQUATH with the Feynman method

The paper does not specify the exact algorithm or class of distributions $\mathcal{D}$ for which such algorithm fails to refute XQUATH, and some classes of distributions $\mathcal{D}$ do not satisfy ...
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  • 326
3 votes
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Quantum hardness of XQUATH conjecture

Maybe think of it this way - a quantum computer, executing a small enough random circuit $C$ acting on a state initially prepared as $\vert 0^n\rangle$ and sampling therefrom, will get an $n$-bit ...
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  • 7,444
3 votes
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Quantum supremacy: shallow depth Haar random circuits and unitary designs

First of all, that does not imply anything for shorter (constant/logarithmic) depths. Moreover, the 2-design property does not imply that the outcome distribution is the same as for Haar-random ...
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3 votes

What is the probability $\Pr(\|U-I\|_{\rm op}<\varepsilon)$ of a Haar-random unitary being close to the identity?

$U-I$ is a normal matrix so $||U-I||_{op}$ is its eigenvalue with the largest magnitude. The eigenvalue equation for this matrix is $$(U-I)|\psi\rangle=\lambda|\psi\rangle,$$ so $$|\lambda|^2=(\cos\...
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3 votes

On the distribution of the fidelity of a random product state with an arbitrary many-qubit state

All you need are simple tools from measure concentration. The setup is as follows (repeated from the question above for completeness): $| \psi \rangle$ is an $n$-qubit state and $| \alpha \rangle := | ...
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2 votes

On the distribution of the fidelity of a random product state with an arbitrary many-qubit state

The required fidelity $F$ is a function of the Cartesian product of the single $n$-qubit state space: $CP^{2^n-1}$ and $n$ copies of a single qubit state space: $CP^{1} \cong S^2$. The statistics ...
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2 votes
Accepted

Random quantum circuits and general efficient POVM measurement

In the absence of additional assumptions, $\mathbb{E}[p_i]$ can be any real number in $[0, 1]$. For example, let $a\in[0,1]$ and define the POVM as $M_0=aI$ and $M_1=(1-a)I$. Then $$ \mathbb{E}[p_0] = ...
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2 votes
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Average output state of random quantum circuits

Calculating $\rho_1$ Let $N=2^n$ denote the dimension of the Hilbert space where $|\psi\rangle$ lives. For $i=0,\dots,N-1$, let $V_i$ be any unitary that maps $|i\rangle$ to $|0\rangle$. The action of ...
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  • 14.4k
1 vote

Sampling Haar over two systems

I would choose to think of $|\psi\rangle$ as $U|0\rangle$ where $U$ is any unitary. But, I can also think of it as $U'|1\rangle$, or $U''|2\rangle,\ldots$. Hence, I can write this as \begin{align*} \...
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  • 47.5k
1 vote

Question regarding integration of Haar random state

I was also flummoxed by the apparent puzzle why the value of $\mathbb{E}[\langle z|C|0^n\rangle|^2]$ is $2/2^n$ instead of $1/2^n$, but in my opinion I think the confusion arises from the symbol $\...
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